A firefighter with a weight of 707 N slides down a vertical pole with an acceleration of 3.04 m/s2, directed downward. (a) What are the magnitude and direction of the vertical force (use up as the positive direction) exerted by the pole on the firefighter?

Fup=492.072N

Explanation:

Given                         required                   solution.

W=707N                     Fup=?            from newton's 2nd law we know that

let g=10N/kg                                  Fnet=maeq1      

a=3.04m/s2                                   so the net force is the total of all      

                                             forces.In other words  

                                      Fnet=W-Fupeq2

substitute eq1 from eq2 so it becomes

ma=w-Fupsubstitute the given numbers

before that we should have to find the mass of the fighter to compute Fup by using the following formula.

w=mg   m=w/g⇒707/10=70.7kg

then continue with the above equation.

ma=w-Fup       use the negative sign in front of Fup because it acts opposite

ma-w=-Fup         take w to the left to get Fup

w-ma=Fup           divide both sides by negative one

707N-70.7kg×3.04m/s2=Fup

Fup= 707N-214.928N

Fup=492.072N

Note that the following;

W is weight of fighter

Fup is the upward reaction force.

m  is the mass of the fighter

a is the acceleration of the fighter.

explanation:

as the height from which the ruler was dropped increased, the speed of the ruler at ryan's hand increased.

grav forces affect all objects magnetic forces affect magnetic objects and electric currents